The shape of the lesson
By the end of the lesson, learners can
- A vector is a journey: In the lab, drag the head and the tail of the vector.
- Adding and subtracting: In the lab, place \(\mathbf{b}\) so its tail sits on the head of \(\mathbf{a}\).
- Multiplying by a scalar: In the lab, multiply a vector by 2, by \(\frac{1}{2}\) and by \(-1\).
- Between points: \(\overrightarrow{AB}\) = end minus start.
- Add or subtract: combine the top numbers, then the bottom numbers.
- Scalar: \(k\mathbf{a}\) multiplies both numbers by \(k\).
- Parallel: one vector is a number times the other.
- Unknowns: make one equation from the tops and one from the bottoms.
One picture every idea returns to
A column vector \(\begin{pmatrix}x\\y\end{pmatrix}\) means \(x\) across and \(y\) up. Add or subtract vectors by combining the top numbers and the bottom numbers separately; \(k\mathbf{a}\) multiplies both numbers by \(k\) and is parallel to \(\mathbf{a}\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Size and direction.. Extended. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: A vector is a journey | In the lab, drag the head and the tail of the vector. Project tab 1 of the lab (Vectors lab, column vectors) and let a student drive it while the class predicts. Model the worked example Vector between two points on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Adding and subtracting | In the lab, place \(\mathbf{b}\) so its tail sits on the head of \(\mathbf{a}\). Project tab 2 of the lab (Vectors lab, add and subtract) and let a student drive it while the class predicts. Model the worked example Sum and difference on the board, one line at a time. Your turn: \(\mathbf{u} = \begin{pmatrix}-1\\4\end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix}3\\-2\end{pmatrix}\). Find \(\mathbf{u} + \mathbf{v}\) and \(\mathbf{v} - \mathbf{u}\). | Pairs |
| 21 to 30 min | Teach and check: Multiplying by a scalar | In the lab, multiply a vector by 2, by \(\frac{1}{2}\) and by \(-1\). Project tab 3 of the lab (Vectors lab, scalar multiples) and let a student drive it while the class predicts. Model the worked example A combination of vectors on the board, one line at a time. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Multiplying by a scalar, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Writing 'start minus end'. \(\overrightarrow{AB}\) is \(B - A\): the journey FROM \(A\) TO \(B\). | Getting it backwards flips both signs and gives \(\overrightarrow{BA}\). |
| Sign slips when subtracting a negative: \(5 - (-3) = 8\), not 2. | Write the subtraction out in full for each number before you simplify. |
| Multiplying only the top number. \(3\begin{pmatrix}2\\k\end{pmatrix}\) is \(\begin{pmatrix}6\\3k\end{pmatrix}\): the scalar multiplies every component. | Return to Multiplying by a scalar: ask a pair to state the correct version and show it on the lab before moving on. |
Vectors in two dimensions: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
\(A\) is the point \((-3, 2)\) and \(B\) is the point \((1, -4)\). Write as column vectors (a) \(\overrightarrow{AB}\) (b) \(\overrightarrow{BA}\).
\(\mathbf{a} = \begin{pmatrix}5\\-2\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-1\\3\end{pmatrix}\). Find (a) \(\mathbf{a} + \mathbf{b}\) (b) \(\mathbf{a} - \mathbf{b}\).
With \(\mathbf{a}\) and \(\mathbf{b}\) as in the previous question, find \(3\mathbf{a} - 2\mathbf{b}\).
Find \(p\) and \(q\) when \(p\begin{pmatrix}3\\-1\end{pmatrix} + q\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}7\\7\end{pmatrix}\).
Show that the vectors \(\begin{pmatrix}10\\-4\end{pmatrix}\) and \(\begin{pmatrix}-15\\6\end{pmatrix}\) are parallel, and state whether they point the same way.
A boat would move by \(\begin{pmatrix}5\\2\end{pmatrix}\) km each hour in still water. A current moves it by \(\begin{pmatrix}-1\\2\end{pmatrix}\) km each hour. The boat starts at the origin.
(a) Find the boat's actual movement each hour as a column vector.
(b) Find its position after 3 hours.
(c) How many hours does it take to reach the buoy at \((20, 20)\)?
Answer key and marking notes
- (a) \(\begin{pmatrix}1 - (-3)\\-4 - 2\end{pmatrix} = \begin{pmatrix}4\\-6\end{pmatrix}\)
- (b) \(\begin{pmatrix}-4\\6\end{pmatrix}\)
- (a) \(\begin{pmatrix}4\\1\end{pmatrix}\)
- (b) \(\begin{pmatrix}5 - (-1)\\-2 - 3\end{pmatrix} = \begin{pmatrix}6\\-5\end{pmatrix}\)
- \(3\mathbf{a} = \begin{pmatrix}15\\-6\end{pmatrix}\), \(2\mathbf{b} = \begin{pmatrix}-2\\6\end{pmatrix}\).
- \(3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix}17\\-12\end{pmatrix}\).
- Tops: \(3p + q = 7\). Bottoms: \(-p + 2q = 7\).
- From the first, \(q = 7 - 3p\). Substitute: \(-p + 14 - 6p = 7\), so \(7p = 7\) and \(p = 1\).
- \(q = 7 - 3 = 4\). Check: \(\begin{pmatrix}3 + 4\\-1 + 8\end{pmatrix} = \begin{pmatrix}7\\7\end{pmatrix}\).
- \(-15 = -1.5 \times 10\) and \(6 = -1.5 \times (-4)\), so \(\begin{pmatrix}-15\\6\end{pmatrix} = -1.5\begin{pmatrix}10\\-4\end{pmatrix}\).
- They are parallel; the multiplier is negative, so they point in opposite directions.
- (a) \(\begin{pmatrix}5\\2\end{pmatrix} + \begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}4\\4\end{pmatrix}\).
- (b) \(3\begin{pmatrix}4\\4\end{pmatrix} = \begin{pmatrix}12\\12\end{pmatrix}\), so it is at \((12, 12)\).
- (c) \(\begin{pmatrix}20\\20\end{pmatrix} = 5\begin{pmatrix}4\\4\end{pmatrix}\), so 5 hours.
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The vector \(\mathbf{a}\) is drawn on the square grid. Write \(\mathbf{a}\) as a column vector.
- A\(\begin{pmatrix}5\\-2\end{pmatrix}\)
- B\(\begin{pmatrix}-5\\2\end{pmatrix}\)
- C\(\begin{pmatrix}5\\2\end{pmatrix}\)
- D\(\begin{pmatrix}-2\\5\end{pmatrix}\)
\(\mathbf{a} = \begin{pmatrix}4\\-3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-2\\5\end{pmatrix}\). Find \(\mathbf{a} - \mathbf{b}\).
- A\(\begin{pmatrix}6\\-8\end{pmatrix}\)
- B\(\begin{pmatrix}2\\2\end{pmatrix}\)
- C\(\begin{pmatrix}6\\2\end{pmatrix}\)
- D\(\begin{pmatrix}2\\-8\end{pmatrix}\)
\(\mathbf{a} = \begin{pmatrix}2\\-1\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-3\\4\end{pmatrix}\). Find \(3\mathbf{a} + 2\mathbf{b}\).
- A\(\begin{pmatrix}0\\5\end{pmatrix}\)
- B\(\begin{pmatrix}0\\11\end{pmatrix}\)
- C\(\begin{pmatrix}0\\3\end{pmatrix}\)
- D\(\begin{pmatrix}-1\\3\end{pmatrix}\)
\(P\) is the point \((3, -1)\) and \(\overrightarrow{PQ} = \begin{pmatrix}-4\\6\end{pmatrix}\). Find the coordinates of \(Q\).
- A\((-1, 5)\)
- B\((7, -7)\)
- C\((-4, 6)\)
- D\((-1, -7)\)
Find \(k\) and \(m\) when \(k\begin{pmatrix}3\\2\end{pmatrix} + \begin{pmatrix}m\\-1\end{pmatrix} = \begin{pmatrix}4\\9\end{pmatrix}\).
- A\(k = 5\), \(m = -11\)
- B\(k = 4\), \(m = -8\)
- C\(k = 5\), \(m = 11\)
- D\(k = 5\), \(m = 19\)
The vector \(\begin{pmatrix}k\\-6\end{pmatrix}\) is parallel to \(\begin{pmatrix}-2\\3\end{pmatrix}\). Find \(k\).
- A\(4\)
- B\(-4\)
- C\(9\)
- D\(-11\)
Quiz key
From the tail to the head the arrow goes 5 squares right and 2 squares DOWN, so \(\mathbf{a} = \begin{pmatrix}5\\-2\end{pmatrix}\). The tempting \(\begin{pmatrix}5\\2\end{pmatrix}\) forgets that going down is negative. \(\begin{pmatrix}-5\\2\end{pmatrix}\) reads the arrow from head to tail (that is \(-\mathbf{a}\)), and \(\begin{pmatrix}-2\\5\end{pmatrix}\) puts the components in the wrong order.
Top: \(4 - (-2) = 6\). Bottom: \(-3 - 5 = -8\). So \(\mathbf{a} - \mathbf{b} = \begin{pmatrix}6\\-8\end{pmatrix}\). The tempting \(\begin{pmatrix}2\\2\end{pmatrix}\) is \(\mathbf{a} + \mathbf{b}\). \(\begin{pmatrix}6\\2\end{pmatrix}\) gets \(-3 - 5\) wrong as \(+2\), and \(\begin{pmatrix}2\\-8\end{pmatrix}\) gets \(4 - (-2)\) wrong as 2.
\(3\mathbf{a} = \begin{pmatrix}6\\-3\end{pmatrix}\) and \(2\mathbf{b} = \begin{pmatrix}-6\\8\end{pmatrix}\), so \(3\mathbf{a} + 2\mathbf{b} = \begin{pmatrix}0\\5\end{pmatrix}\). The tempting \(\begin{pmatrix}0\\11\end{pmatrix}\) loses the sign in \(3 \times (-1)\), using \(+3\). \(\begin{pmatrix}0\\3\end{pmatrix}\) multiplies only the top numbers by the scalars, and \(\begin{pmatrix}-1\\3\end{pmatrix}\) is just \(\mathbf{a} + \mathbf{b}\).
\(Q = (3 + (-4), -1 + 6) = (-1, 5)\). The tempting \((7, -7)\) subtracts the vector, which is the journey from \(Q\) back to \(P\) the wrong way round. \((-4, 6)\) writes the vector itself as the point, forgetting the start at \(P\), and \((-1, -7)\) adds across correctly but subtracts the 6.
Bottom: \(2k - 1 = 9\), so \(2k = 10\) and \(k = 5\). Top: \(3k + m = 4\), so \(15 + m = 4\) and \(m = -11\). The tempting \(k = 4\) moves the \(-1\) across without changing its sign (\(2k = 8\)), which then makes \(m = 4 - 12 = -8\). \(m = 11\) works out \(15 - 4\) instead of \(4 - 15\), and \(m = 19\) adds 15 to 4.
\(-6 = -2 \times 3\), so the vector is \(-2\begin{pmatrix}-2\\3\end{pmatrix} = \begin{pmatrix}4\\-6\end{pmatrix}\) and \(k = 4\). The tempting \(-4\) multiplies the top by \(+2\) and the bottom by \(-2\); the SAME scalar must multiply both components. 9 sets up the ratio crosswise (\(\dfrac{k}{-6} = \dfrac{3}{-2}\)), and \(-11\) subtracts 9 from \(-2\) because the bottom fell by 9; parallel vectors are multiples, not differences.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.